1
$\begingroup$

I'm trying to solve this problem: Let $\alpha>0$ and $a_{n}=\frac{\cos n}{2n^{\alpha}}$ for all $n\in\mathbb{N}$. Prove that the series $\sum_{n=1}^{\infty}a_{n}$ coneverges.

For $\alpha>1$, we have that $\mid\cos x\mid\leq1$ for all $x\in\mathbb{R}$ and therefore for all $n\in\mathbb{N}$: $$ \left\lvert \frac{\cos n}{2n^{\alpha}}\right\rvert=\frac{\lvert\cos n\rvert}{2n^{\alpha}}\leq\frac{1}{2}\cdot\frac{1}{n^{\alpha}} $$

Using the fact that $\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}$ is convergent for $\alpha>1$, we have for the comparison test that $\sum_{n=1}^{\infty}a_{n}$ is absolutely convergent for $\alpha>1$. And with this $\sum_{n=1}^{\infty}a_{n}$ is convergent. But I don't know how can I prove that $\sum_{n=1}^{\infty}a_{n}$ converges when $0<\alpha<1$. Could you help me or give me some idea to prove this?

$\endgroup$
  • $\begingroup$ Are you sure it converges? $\endgroup$ – herb steinberg Aug 19 '19 at 19:58
  • $\begingroup$ Use an Abel summation, knowing that the partial sums of $\cos{n}$ are bounded. $\endgroup$ – Mindlack Aug 19 '19 at 19:59
2
$\begingroup$

In the following, we derive Abel's test and Dirichlet's test for testing the convergence of infinite series in the form $\sum_{n=1}^{\infty}a_{n}b_{n}$. Firstly, we introduce Abel's transform, which is the discrete version of integration-by-part. Then we apply it to the difference of partial sums $S_{n+k}-S_{n}$.

Section 1: Abel's transform. Let $(a_{n})$ and $(b_{n})$ be sequences of real numbers. For each $n\in\mathbb{N}$, let $B_{n}=\sum_{k=1}^{n}b_{k}$. Define $B_{0}=0$. Let $N\in\mathbb{N}$. We rewrite the partial sum $\sum_{n=1}^{N}a_{n}b_{n}$ as \begin{eqnarray*} \sum_{n=1}^{N}a_{n}b_{n} & = & \sum_{n=1}^{N}a_{n}(B_{n}-B_{n-1})\\ & = & \sum_{n=1}^{N}a_{n}B_{n}-\sum_{n=1}^{N}a_{n}B_{n-1}\\ & = & \sum_{n=1}^{N}a_{n}B_{n}-\sum_{n=1}^{N-1}a_{n+1}B_{n}\\ & = & a_{N}B_{N}-\sum_{n=1}^{N-1}(a_{n+1}-a_{n})B_{n}. \end{eqnarray*} This is known as Abel's transform. Comparing to the well-known integration-by-part formula $\int_{a}^{b}f(x)g'(x)dx=[f(x)g(x)]_{a}^{b}-\int_{a}^{b}g(x)f'(x)dx,$ we note that $(a_{n})$ is analog to $f$ while $(b_{n})$ is analog to $g'$.

Section 2: Define partial sum $S_{n}=\sum_{k=1}^{n}a_{k}b_{k}$. Let $n,k\in\mathbb{N}$. We consider $S_{n+k}-S_{n}$. Applying Abel's transform, we have \begin{eqnarray*} S_{n+k}-S_{n} & = & \sum_{i=n+1}^{n+k}a_{i}b_{i}\\ & = & a_{n+k}(B_{n+k}-B_{n})-\sum_{i=n+1}^{n+k-1}(a_{i+1}-a_{i})(B_{i}-B_{n}). \end{eqnarray*} Now, we are ready to state and prove the following two theorems.

Theorem 1: If $(a_{n})$ is monotone and bounded, and the series $\sum_{n=1}^{\infty}b_{n}$ is convergent, then the series $\sum_{n=1}^{\infty}a_{n}b_{n}$ is convergent.

Proof: Choose $M>0$ such that $|a_{n}|\leq M$ for all $n$. Let $\varepsilon>0$ be arbitrary. Choose $N$ such that $|B_{n+k}-B_{n}|<\varepsilon$ whenever $n\geq N$ and $k\in\mathbb{N}$. For any $n\geq N$ and $k\in\mathbb{N}$, we have that: \begin{eqnarray*} |S_{n+k}-S_{n}| & \leq & |a_{n+k}(B_{n+k}-B_{n})|+\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}||B_{i}-B_{n}|\\ & \leq & M\varepsilon+\sum_{i=n+1}^{n+k-1}\varepsilon|a_{i+1}-a_{i}|\\ & \leq & M\varepsilon+\varepsilon|\sum_{i=n+1}^{n+k-1}(a_{i+1}-a_{i})|\\ & \leq & M\varepsilon+\varepsilon|a_{n+k}-a_{n+1}|\\ & \leq & 3M\varepsilon \end{eqnarray*} This shows that $(S_{n})$ is a Cauchy sequence and hence $\sum_{n=1}^{\infty}a_{n}b_{n}$ is convergent. In the above, we have used the fact that $(a_{n})$ is monotone to obtain $\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}|=|\sum_{i=n+1}^{n+k-1}(a_{i+1}-a_{i})|$.

Theorem 2: If $(a_{n})$ is monotone and $a_{n}\rightarrow0$, and $(B_{n})$ is bounded (where $B_{n}:=\sum_{k=1}^{n}b_{k}$), then the series $\sum_{n=1}^{\infty}a_{n}b_{n}$ is convergent.

Proof: Let $M>0$ be such that $|B_{n}|\leq M$ for all $n$. Let $\varepsilon>0$ be arbitrary. Choose $N$ such that $|a_{n}|<\varepsilon$ whenever $n\geq N$. For any $n\geq N$ and $k\in\mathbb{N}$, we have: \begin{eqnarray*} |S_{n+k}-S_{n}| & \leq & |a_{n+k}(B_{n+k}-B_{n})|+\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}||B_{i}-B_{n}|\\ & \leq & 2M\varepsilon+2M\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}|\\ & = & 2M\varepsilon+2M|a_{n+k}-a_{n+1}|\\ & \leq & 6M\varepsilon. \end{eqnarray*} Therefore $(S_{n})$ is a Cauchy sequence and hence the series $\sum_{n=1}^{\infty}a_{n}b_{n}$ is convergent.

//////////////////////////////////////////////////////////////////////////////////////

Now, we go to prove that for each $\alpha\in(0,1)$, the series $\sum_{n=1}^{\infty}\frac{\cos n}{2n^{\alpha}}$ is convergent.

Proof: Let $a_{n}=\frac{1}{2n^{\alpha}}$, $b_{n}=\cos n$. Clearly $(a_{n})$ is monotonic decreasing and $a_{n}\rightarrow0$. Let $B_{n}=\sum_{k=1}^{n}b_{k}$. We go to show that $(B_{n})$ is bounded, then apply Theorem 2...

We go to compute $\sum_{k=0}^{n-1}\cos k\theta$ and demonstrate the use of complex numbers. Let $\theta\in\mathbb{R}$ be such that $\cos\theta\neq1$. Define $z=\cos\theta+i\sin\theta$. By DeMoivre Theorem, $z^{k}=\cos(k\theta)+i\sin(k\theta)$ for any $k\in\mathbb{N}$. On one hand, \begin{eqnarray*} 1+z+z^{2}+\ldots+z^{n-1} & = & \frac{1-z^{n}}{1-z}\\ & = & \frac{1-(\cos n\theta+i\sin n\theta)}{1-(\cos\theta+i\sin\theta)}\\ & = & \frac{2\sin^{2}(\frac{n\theta}{2})-2i\sin(\frac{n\theta}{2})\cos(\frac{n\theta}{2})}{2\sin^{2}(\frac{\theta}{2})-2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}\\ & = & \frac{\sin(\frac{n\theta}{2})\left[\sin(\frac{n\theta}{2})-i\cos(\frac{n\theta}{2})\right]}{\sin(\frac{\theta}{2})\left[\sin(\frac{\theta}{2})-i\cos(\frac{\theta}{2})\right]}\\ & = & \frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cdot\frac{\left[\sin(\frac{n\theta}{2})-i\cos(\frac{n\theta}{2})\right](i)}{\left[\sin(\frac{\theta}{2})-i\cos(\frac{\theta}{2})\right](i)}\\ & = & \frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cdot\frac{\cos(\frac{n\theta}{2})+i\sin(\frac{n\theta}{2})}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}\\ & = & \frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cdot\left[\cos(\frac{n\theta}{2}-\frac{\theta}{2})+i\sin(\frac{n\theta}{2}-\frac{\theta}{2})\right]. \end{eqnarray*} On the other hand, \begin{eqnarray*} 1+z+z^{2}+\ldots+z^{n-1} & = & \sum_{k=0}^{n-1}\cos k\theta+i\sum_{k=1}^{n-1}\sin k\theta. \end{eqnarray*} Comparing the real part, we obtain: $$ \sum_{k=0}^{n-1}\cos k\theta=\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cos\left(\frac{(n-1)\theta}{2}\right). $$ If $\cos\theta=1$, then $\theta=2m\pi$ for some $m\in\mathbb{Z}$. It follows that $\cos k\theta=1$ for all $k\in\mathbb{N}$. Therefore $\sum_{k=0}^{n-1}\cos k\theta=n.$ We conclude that: If $\theta\in\mathbb{R}$ such that $\cos\theta\neq1$, then $$ |\sum_{k=0}^{n-1}\cos k\theta|\leq\left|\frac{1}{\sin(\frac{\theta}{2})}\right|. $$ Hence, the partial sum of the infinite series $1+\cos\theta+\cos2\theta+\ldots$ is bounded by $\left|\frac{1}{\sin(\frac{\theta}{2})}\right|$. Clearly $\cos1\neq1$, so $(B_{n})$ is bounded.

$\endgroup$
3
$\begingroup$

Hint

Dirichlet’s test may be your friend.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.